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Math Help - Linear Algebra please help.

  1. #1
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    Linear Algebra please help.

    T:R2-->R3 defined by T(a1,a2)=(a1+a2,0,2a1-a2)

    I need to prove that is a linear transformation, and find the bases for both
    N(T) and R(T). then compute the nullity and rank of T. finally say where it is one -2- one or onto.

    I already got the anwsers but I need to know in detail especially for computing the nullity and rank of T. and onto or one 2 one.
    Thanks
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by JCIR View Post
    T:R2-->R3 defined by T(a1,a2)=(a1+a2,0,2a1-a2)

    I need to prove that is a linear transformation
    Its just checking rules right? Do it.

    Or get a matrix that represents the transformation.

    T \equiv \begin{pmatrix}1 & 1 \\ 0 & 0 \\ 2 & -1 \end{pmatrix}

    and find the bases for both
    Consider the standard base for \mathbb{R}^2 = \{(0,1),(1,0)\}.

    Now T(0,1) = (1,0,-1),T(1,0) = (1,0,2)

    So the basis for T(\mathbb{R}^2).

    N(T) and R(T). then compute the nullity and rank of T. finally say where it is one -2- one or onto.
    N(T) is (x,y):T(x,y) = (x+y,0,2x - y) = 0 \Rightarrow -x = y = 2x \Rightarrow (x,y) = 0..This means Nullity = 0

    R(T) is (a,b,c):T(x,y) = (x+y,0,2x - y) = (a,b,c) \Rightarrow y = a-x, 2x-y = 3x - a = c  \Rightarrow (x,y) = (\frac{a+c}3,\frac{2a- c}3 )..

    This means R(T) will be the subspace of R3, in which the second co-ordinate is 0.This means Rank = 2.

    The map is one-one because nullity is 0.
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